smart hendsome
2016-Feb-04 04:08 UTC
[R] Generate arrival of time based on uniform random number
Hi everyone, I have problem regarding to generate arrival of time based on uniform random number. Let day0 = 0.383, lambda = 0.2612 1) Generate uniform random number (already settled) using the code below: set.seed(1234) rand.no <- function(n,itr){ ? matrix(runif(n*itr, 0, 1), nrow=n, ncol=itr) } x<-rand.no(10,1) 2) For day 1, ?? a) if 1st rand.no < day0, then its not rain,? else its rain (i want the rseult in next colum, let say rain) 3) For day 2 and so on ( it depend from next rain when it happened) 4) Next rain, ?? to know the next rain i need i need to calculate using the exponential ? a) 1st rain = - ln (1st rand.no) / lambda ? b) 2nd nextRain = 1st rain - ln (2nd rand.no) / lambda ? c) 3rd nextRain = 2nd next rain - ln (3rd rand.no)/lambda,? and so on ?eg: if nextRain = 2.435, then 2nd day will rain Hope anyone can help me solve this problems. Thanks so much. ????? ? [[alternative HTML version deleted]]
Rolf Turner
2016-Feb-04 04:17 UTC
[R] [FORGED] Generate arrival of time based on uniform random number
This looks very much like a homework problem and this list has a "no homework" policy. cheers, Rolf Turner -- Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 On 04/02/16 17:08, smart hendsome via R-help wrote:> Hi everyone, > I have problem regarding to generate arrival of time based on uniform random number. > Let day0 = 0.383, lambda = 0.2612 > > 1) Generate uniform random number (already settled) using the code below: > set.seed(1234) > rand.no <- function(n,itr){ > matrix(runif(n*itr, 0, 1), nrow=n, ncol=itr) > } > x<-rand.no(10,1) > 2) For day 1, > a) if 1st rand.no < day0, then its not rain, else its rain (i want the rseult in next colum, let say rain) > 3) For day 2 and so on ( it depend from next rain when it happened) > 4) Next rain, > to know the next rain i need i need to calculate using the exponential > > a) 1st rain = - ln (1st rand.no) / lambda > b) 2nd nextRain = 1st rain - ln (2nd rand.no) / lambda > c) 3rd nextRain = 2nd next rain - ln (3rd rand.no)/lambda, and so on > eg: if nextRain = 2.435, then 2nd day will rain > > > Hope anyone can help me solve this problems. Thanks so much.